Question

The area of a parking lot is 1710 square meters. A car requires 5 square meters and a bus requires 32 square meters of space. There can be at most 180 vehicles parked at one time. If the cost to park a car is $2.00 and a bus is $6.00, how many buses should be in the lot to maximize income? Please help :(

Answer 1

Givens

• The area of the parking lot is 1710 square meters.
• A car requires 5 square meters.
• A bus requires 32 square meters.
• There can be a maximum of 180 vehicles.
• The cost for a car is $2.00.
• The cost for a bus is $6.00.

To solve this problem we need to create a table to order all this information and express it as a system of inequations.

Car Bus Total Capcity

Sq. Meters 5c 32b 1710

N° vehicles c b 180

Therefore, the inequalities are

`5c+32b\leq 1710\\c+b\leq 180`

The expresion which represents the income is

`I=2c+6b`, because a car is $2.00 and a bus is $6.00.

Now, we first need to find the critical points of the solution of the inequality system, which is attached. Observe that the only points that can be a solution is (150,30), because there can't be just car or just buses.

Then, we replace this point in the income expression

`I=2c+6b\\I=2(150)+6(30)=300+180=480`

Therefore, in order to maximize incomes, we need to park 150 cars and 30 buses, to make $480 income.

Answer 2

To maximize the income should be 30 buses and 150 cars

Explanation:

Let

x-----> the number of cars

y ----> the number of bus

we know that

`5x+32y\leq1,710` ------> inequality A

`x+y\leq 180` ----> inequality B

The function of the cost to maximize is equal to

`C=2x+6y`

Solve the system of inequalities by graphing

The solution is the shaded area

see the attached figure

The vertices of the solution are

(0,0),(0,53),(150,30),(180,0)

Verify

(0,53) --->
`C=2(0)+6(53)=\$318`

(150,30) --->
`C=2(150)+6(30)=\$480`

therefore

To maximize the income should be 30 buses and 150 cars